Let A = 3 − t − 1 0 1 3 − t − 1 0 1 0 and det A = 5 , then, t = 1, t = 2, t = − 1, t = − 2

Hint: ∣ A ∣ = 3 − t − 1 0 1 3 − t − 1 0 1 0 = 3 − t = 5 , t = − 2

Let A = 3 − t − 1 0 1 3 − t − 1 0 1 0 and det A = 5 , then

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Let $\mathrm{A}=\left(\begin{array}{ccc}3-\mathrm{t} & 1 & 0 \\ -1 & 3-\mathrm{t} & 1 \\ 0 & -1 & 0\end{array}\right)$ and $\operatorname{det} \mathrm{A}=5$, then

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Options:

A $\mathrm{t}=1$
B $\mathrm{t}=2$
C $t=-1$
D $\mathrm{t}=-2$

Solution:

2970 Upvotes Verified Answer

The correct answer is: $\mathrm{t}=-2$

Hint:
$|\mathrm{A}|=\left|\begin{array}{ccc}3-\mathrm{t} & 1 & 0 \\ -1 & 3-\mathrm{t} & 1 \\ 0 & -1 & 0\end{array}\right|=3-\mathrm{t}=5, \mathrm{t}=-2$

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